Apply the length contraction formula $L = L_0 / \gamma$ and the relativistic momentum formula $p = \gamma m v$ to predict the contraction of moving objects and the momentum of relativistic particles

What this dot point is asking

QCAA wants you to apply the length contraction and relativistic momentum formulas in numerical problems, recognising that both effects scale by the Lorentz factor $\gamma$ at high speeds.

Length contraction

A rod at rest in some frame has the proper length $L_0$, measured by an observer in that frame. From a different frame moving relative to the rod at speed $v$, the rod is shorter along the direction of motion:

where $\gamma = 1/\sqrt{1 - v^2/c^2}$.

Direction matters

Length contraction acts only along the direction of motion. A rod's transverse dimensions (perpendicular to its motion) are unaffected. A passing ball would be seen as flattened along its direction of motion but unchanged in the other two dimensions.

Proper length

The proper length is the longer length, measured in the frame where the rod is at rest. Other observers measure a shorter length.

Length contraction is real

Like time dilation, length contraction is not optical illusion. The atmosphere appears thinner to a relativistic muon (in the muon's frame, the atmosphere is contracted; in Earth's frame, time is dilated; both descriptions agree on whether the muon survives to the surface).

Worked example

A spaceship has proper length $L_0 = 100$ m. It travels at $v = 0.80 c$ relative to Earth.

$\gamma = 1/\sqrt{1 - 0.64} = 1/\sqrt{0.36} = 1/0.6 \approx 1.667$.

$L = 100 / 1.667 \approx 60$ m.

From Earth, the spaceship is 60 m long.

Relativistic momentum

In classical mechanics, momentum is $p = m v$. At relativistic speeds, this formula breaks down. The relativistic momentum is:

where $\gamma = 1/\sqrt{1 - v^2/c^2}$ and $m$ is the rest mass.

Why the modification

Classical momentum is not conserved under Lorentz transformations between frames. Relativistic momentum is conserved. Several derivations are possible; the simplest is to require that momentum conservation hold in all frames, which forces the $\gamma$ factor.

Behaviour at high speeds

At low speeds, $\gamma \approx 1$ and $p \approx m v$ (classical). At high speeds, $\gamma$ grows without bound as $v \to c$. The momentum required to push a particle to a speed approaching $c$ grows without limit; no particle with rest mass can reach $c$.

Worked example

An electron is accelerated to $v = 0.99 c$. Find its momentum.

$\gamma = 7.09$ (from above).

$p = 7.09 \times (9.11 \times 10^{-31}) \times (0.99 \times 3 \times 10^8) = 7.09 \times 2.71 \times 10^{-22} \approx 1.92 \times 10^{-21}$ kg m s$^{-1}$.

The classical value would be $2.71 \times 10^{-22}$, smaller by a factor of $\gamma = 7.09$.

Consequences of length contraction and relativistic momentum

Particle accelerators

Designing a particle accelerator requires the relativistic momentum formula. Synchrotrons (where charged particles travel in a circle in a magnetic field) must use $p = \gamma m v$ when calculating the magnetic field needed to keep particles in their circular path.

Cosmic rays

Some high-energy cosmic ray particles have $\gamma > 10^{10}$ (extraordinarily relativistic). The classical momentum formula is hopelessly wrong; only $p = \gamma m v$ works.

No object with rest mass reaches $c$

The relativistic energy $E = \gamma m c^2$ (covered in the mass-energy dot point) diverges as $v \to c$. Infinite energy would be required.

Photons

Photons have zero rest mass and travel at $c$. They carry momentum $p = E/c$ (consistent with $p = \gamma m v$ in an appropriate limit).

Verification

Particle accelerator data. Charged particles in accelerators (LHC, Tevatron, etc.) follow trajectories that match relativistic momentum predictions. Without the $\gamma$ correction, the accelerator's magnetic systems would not bend the particles correctly.

Pion decay. High-energy pions in cosmic-ray cascades have lifetimes consistent with time dilation, and trajectories consistent with relativistic momentum.

Examples in context

Example 1. A cosmic-ray muon traversing the atmosphere at $v = 0.99c$ has $\gamma = 1/\sqrt{1-0.99^2} = 7.09$. Earth-frame muon-travel distance of $L_0 = 10 \text{ km}$ contracts in the muon's rest frame to $L = L_0/\gamma = 1.41 \text{ km}$ - short enough for the muon to traverse before decay. ANSTO Mt Cotton students analysing cosmic-ray data apply $L = L_0/\gamma$ directly.

Example 2. A $0.95c$ particle of rest mass $m_0 = 1.0 \times 10^{-27} \text{ kg}$ at a synchrotron facility has $\gamma = 1/\sqrt{1-0.9025} = 3.20$. Relativistic momentum $p = \gamma m_0 v = 3.20 \times 10^{-27} \times 2.85 \times 10^8 = 9.13 \times 10^{-19} \text{ kg m s}^{-1}$, far exceeding the classical $m_0 v = 2.85 \times 10^{-19}$. QCAA EA Unit 4 thematic items contrast these explicitly.

Try this

Q1. State the length-contraction formula and define proper length. [2 marks]

• Cue. $L = L_0/\gamma$; $L_0$ measured in the object's rest frame.

Q2. Calculate the contracted length of a $30 \text{ m}$ rod at $v = 0.80c$. [3 marks]

• Cue. $\gamma = 1/\sqrt{1-0.64} = 1.667$; $L = 30/1.667 = 18 \text{ m}$.

Q3. A cosmic-ray muon at $0.99c$ tracked over a $10 \text{ km}$ atmospheric column. (a) Calculate $\gamma$ and the contracted length in the muon's frame. (b) Calculate the relativistic momentum of a $1.88 \times 10^{-28} \text{ kg}$ muon. (c) Justify why classical mechanics fails at this speed. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $\gamma = 7.09$, $L = 1.41 \text{ km}$; (b) $p = \gamma m v = 3.95 \times 10^{-19} \text{ kg m s}^{-1}$; (c) classical underpredicts by factor $\gamma$.